Lemma 76.15.9 (060B)—The Stacks project

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Table of contents
Part 4: Algebraic Spaces
Chapter 76: More on Morphisms of Spaces
Section 76.15: Universal first order thickenings
Lemma 76.15.9 (cite)

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Lemma 76.15.9. Let $S$ be a scheme. Let

$\xymatrix{ Z \ar[r]_ h \ar[d]_ f & X \ar[d]^ g \\ W \ar[r]^{h'} & Y }$

be a fibre product diagram of algebraic spaces over $S$ with $h'$ formally unramified. Then $h$ is formally unramified and if $W \subset W'$ is the universal first order thickening of $W$ over $Y$ , then $Z = X \times _ Y W \subset X \times _ Y W'$ is the universal first order thickening of $Z$ over $X$ . In particular the canonical map $f^*\mathcal{C}_{W/Y} \to \mathcal{C}_{Z/X}$ of Lemma 76.15.8 is surjective.

Proof. The morphism $h$ is formally unramified by Lemma 76.14.5. It is clear that $X \times _ Y W'$ is a first order thickening. It is straightforward to check that it has the universal property because $W'$ has the universal property (by mapping properties of fibre products). See Lemma 76.5.5 for why this implies that the map of conormal sheaves is surjective. $\square$

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